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3 LOGARITHMS AND EXPONENTIALS 87

has the same meaning as the algebraic one). In addition, we define 1* to
be 1 for all real numbers x. Then, for a > 0, x,y arbitrary real numbers,
we have by definition a^x+y = a^xa^y, a"^x = l/a^x, a^Q = 1. Replacing x by b^xin (4.3.3) yields

(4.3.4) log_a(b^x) = x log. b (b>0,x real)
and replacing b by a^y in that formula gives

(4.3.5) (a^x)^y = a^xy (x, y real, a > 0).

For a>l₉ x-*a^x is strictly increasing and such that lim a^x = Q,

x~* — oo

lim a^x = +00; for a < 1, x~*a^x is strictly decreasing, and such that

X-+ + 00

lim a^x = +00, lim a^x = 0.

(4.3.6) The mapping (x₉ y)-*x^y is continuous in R* x R, and tends to a
limit at each point o/E x R in the closure o/R* x R and distinct from (0, 0),

1, -00).

From (4.3.4), we have x^y = a^y'^l°^SaX (a fixed number >1), hence the
result by (4.1 .2) and (4.1 .9).

(4.3.7) Any continuous mapping g o/R* into itself such that g(xy) = g(x)g(yj
has the form x -> ;c^fl, with a real.

Indeed, if b > 1, f(x) = log_& g(b^x) is such that f(x + y) =/(*)
for real x, y, and is continuous, hence f(x) = c -x by (4.1.3), therefore
_g(b^x) = b^cx = (b^x)^c, which proves the result.

As Iog₆(;c^a) = a -log_fc x, we see that if a > 0, x-+x^a is strictly increas-
ing, and strictly decreasing if 0<0; moreover if a > 0, lim^:^fl = 0,

x-+Q

lim x^a = + oo ; if a < 0, lim x^a = +00, lim x^a = 0. For a ^ 0,

X-+ + QO x-+Q x-»- + oo

x-+x^a is therefore a homeomorphism of R* onto itself, by (4.2.2); the
inverse homeomorphism is x-+x^l/a.

PROBLEM

Let/be a mapping of R into itself such that f(x + y) —f(x) +/(>>) and/(;x;y) =*f(x)f(y).
Show that either /(*) = 0 for every x e R, or /(*) « A: for every x e R. (If/(I) ^ 0, then
/(I) == 1; in the second case, show that/(;c) = x for rational x, and using the fact that every
real number z > 0 is a square, show that / is strictly increasing.)



	3 LOGARITHMS AND EXPONENTIALS 87 has the same meaning as the algebraic one). In addition, we define 1* to be 1 for all real numbers x. Then, for a > 0, x,y arbitrary real numbers, we have by definition a^x+y = a^xa^y, a"^x = l/a^x, a^Q = 1. Replacing x by b^xin (4.3.3) yields (4.3.4) log_a(b^x) = x log. b (b>0,x real) and replacing b by a^y in that formula gives (4.3.5) (a^x)^y = a^xy (x, y real, a > 0). For a>l₉ x-a^x* is strictly increasing and such that lim a^x = Q, x~* — oo lim a^x = +00; for a < 1, x~a^x* is strictly decreasing, and such that X-+ + 00 lim a^x = +00, lim a^x = 0.

	(4.3.6) The mapping (x₉ y)-x^y is continuous in* R* x R, and tends to a limit at each point o/E x R in the closure o/R* x R and distinct from (0, 0), 1, -00).

	From (4.3.4), we have x^y = a^y'^l°^SaX (a fixed number >1), hence the result by (4.1 .2) and (4.1 .9).

	(4.3.7) Any continuous mapping g o/R* into itself such that g(xy) = g(x)g(yj has the form x -> ;c^fl, with a real.

	Indeed, if b > 1, f(x) = log_& g(b^x) is such that f(x + y) =/() for real x, y,* and is continuous, hence f(x) = c -x by (4.1.3), therefore _g(b^x) = b^cx = (b^x)^c, which proves the result. As Iog₆(;c^a) = a -log_fc x, we see that if a > 0, x-+x^a is strictly increas- ing, and strictly decreasing if 0<0; moreover if a > 0, lim^:^fl = 0, *x-+Q* lim x^a = + oo ; if a < 0, lim x^a = +00, lim x^a = 0. For a ^ 0, X-+ + QO x-+Q x-»- + oo x-+x^a is therefore a homeomorphism of R* onto itself, by (4.2.2); the inverse homeomorphism is x-+x^l/a.

	PROBLEM Let/be a mapping of R into itself such that f(x + y) —f(x) +/(>>) and/(;x;y) =f(x)f(y).* Show that either /() = 0 for every x e* R, or /() « A: for every x e R. (If/(I) ^ 0, then /(I) == 1; in the second case, show that/(;c) = x for rational x, and using the fact that every real number z >* 0 is a square, show that / is strictly increasing.)